A Worked Example in 2-Dimensions
Normal Lattice Space
Starting information (ignoring the z-axis):
Miller’s indices:
Intercepts
Invert
(1, , )
(1,0,0)
(0,1,0)
(, 1, )
(1, 1, )
(1,1,0)
(1, 2, )
(1,½,0)
“Integer-ize”
to get index
(1,0,0)
(0,1,0)
(1,1,0)
(2,1,0)
Reciprocal Lattice Space
Starting information:
Also because it is an elastic scatter,
wavevector and
, where
is the diffracted wavevector.
is the incident
Normal
Reciprocal
Condition for Bragg diffraction peak
Condition for Bragg diffraction peak:
OR
Note: the range of n is limited to integers such that is real.
WHY??
For diffraction from the (1,0,0) planes, the separation d = a = 1
inch and one obtains:
n
n /2a
2
1
0.35
20.5
41
2
0.70
44.4
88.8
3
1.05
One can find all of the scattering angles by construction (Called Ewald
construction) First attach the head of to a lattice point, with its tail free
to rotate, which is related to the crystal rotating in the incident beam. To
the tail of , attach the tail of . Place the head of
on a second lattice
point and measure the angle between and . This is the angle between
the incident beam and the diffracted beam to a peak, and is equal to 2 Recall that is the angle in the standard Bragg equation. The limiting
condition on the integers m and n is that
. (Hint: When
doing the Ewald’s construction, one can work almost exclusively in the
first quadrant.)
Continue these calculations for the (0,1,0) series of planes.
For diffraction from the (1,1,0) planes, the separation
= 0.79 inch and one obtains:
n
n /2d
2
1
0.44
26.3
52.6
2
0.89
62.4
125
3
1.33
Continue these calculations for the (2,1,0), (1,2,0), and (1,3,0)
series of planes.
Do the planes with indices that have larger integers get closer
or further apart?
Is there a maximum or a minimum distance, d, between
reflecting plane beyond which interference will create
“peaks?” If so what is it?
WHY is the angle between
WHY must
?
and
equal to 2 ?
Do the Ewald construction for this example crystal and verify the
angles found on the left using 2d sin = .
So now to calculate the angle of diffraction peaks for a known crystal one can either use the Bragg equation directly or find
the reciprocal lattice and do and Ewald construction. But that is not the problem at hand, which is the inverse problem. How does
one start with the angles of the diffraction peaks and obtain the crystal’s basis vectors? The basis vectors determine the spacing in the
crystal and associated angles. This is often a dark art, but for simple 2D crystals it is not impossible.
After making measurements, in a perfect world, one has a table of angles , 2 , where a diffraction peak was observed. If we
took measurements on our example crystal, we would have something like:
“Measured” Angles for Example Crystal Peaks and Analysis
Peak number
1
2
3
4
5
6
7
8
9
10
2°
31.2
41
52.6
65.2
81.1
88.8
97.2
108
123
125
°
15.6
20.5
26.3
32.6
40.5
44.4
48.6
53.9
61.7
62.4
sin( ) = n/2d
0.27
0.35
0.44
0.54
0.65
0.70
0.75
0.81
0.88
0.89
Analysis of Data for Crystal in Normal Space
d/n = / (2 sin )
1.3
1
0.80
0.65 =
1.3 / 2
0.54
0.5 =
1/2
0.47
0.43 =
1.3 / 3
0.4 =
0.8 / 2
0.39 =
0.8 / 2
Designation
S1
T1
U1
S2
V
T2
W
S3
U2'
U2'’
Miller( not known
before completing
analysis)
010
100
110
010
120
100
210
010
130
110
Analysis of Data for Crystal in Reciprocal Lattice Space
Reciprocal lattice
separation G
Designation
4.85
6.29
7.90
9.70
11.67
12.57
13.47
14.55
15.80
15.98
G1
G2
G3
G4
G5
G6
G7
G8
G9
G10
Normal
In normal space, one can determine the direction normal to
the reflecting planes, because the normal bisects the angle
between the incident beam and the diffracted beam. The angular
data along with the known wavelength of the microwave, allows
one to calculate d/n. In this data, look for multiples, to help
determine n and thus d. It may take several, iterations of guessing
and testing. It is possible the n = 1 peak is blended with other
peaks and not resolved for small diffraction angles.
For each diffraction peak one plane separation, d, and an
angle, . One can draw a series of parallel lines, each a distance d
from the next. The angles at which these lines are drawn take a
bit more thought. If done correctly, the intersection of the various
lines one series with the other, should indicate the location of the
scattering centers. Start by picking the angle of the incident
microwaves, say vertically down, and an origin where one
scatterer is located. Starting with the peak associated with the
largest plane separation ( that is peak # 1, in this example) draw
the first line through the origin and an angle relative to the
incident microwaves.
Reciprocal
In reciprocal lattice space, the trick for converting
scattering angles to lattice structure goes like this. First find the
separations, denoted G, between individual lattice points in the
reciprocal lattice. In the Ewald construction one has an isosceles
with 2 sides of length
with a common angle, . The unequal
side has a length,
and it is straightforward to show that:
First choose a one point or origin, and a direction. (These
may be chosen randomly, but wisdom suggest starting near the
lower left and choosing either a vertical or horizontal direction. )
Your chosen point is one lattice point. A second lattice point is a
distance G1 from the origin in your chosen direction. In this same
direction one can repeat this process to construct a series of lattice
points along a line a distance G1
apart
From each of these points scribe arcs with radii equal to the other
known G’s. The intersections of the arcs MAY indicate a new
lattice point. Note intersections less than G1 away from the first set
of points probably are not lattice points because they would result
in an even shorter G, unless somehow a peak was missed. With
luck, a missed diffraction peak will be apparent For example if
you you have G’s of length 2 and 3 you might want to check to see
if a G of length 1 was missed. A bit of trial and error and a few
iterations and one can get a fair match to the reciprocal lattice.
Normal
Reciprocal
So the first set of parallel lines is easy. A problem arises
when one naively repeats the process for peak # 2 on the same
figure. The problem is that the scattering planes are about 90°
from where they belong relative to the first set of lines. But things
work better working with information from peak # 3.
With a second row of lattice points for the reciprocal
lattice, one has the basis vectors, and . From these the entire
reciprocal lattice can easily be generated. The trick is to get back
to the real lattice.
Basically this is mostly guess work, trying some
combination and then testing to see if everything works. Using
overlays like overhead transparencies may help. If there crystal is
square ( or equal length sided parallelogram), than two different
sets of planes produce peeks at the same angle. Dealing with this
degeneracy may be tricky.
Note by definition
and
, so the angles are easy.